任务 · 算法困难 Python ·

Convex hull

Return the corners of the convex hull of the points in counter-clockwise order, starting from the point with the smallest x (the smallest y among those). Points on an edge of the hull that aren't corners are left out, and duplicate points count once. If all points coincide, return that one point; if they all lie on one line, return its two endpoints, the starting point first. Coordinates are integers with absolute value up to 10**9. Up to 200,000 points.

函数签名

def convex_hull(points: list[list[int]]) -> list[list[int]]:

示例

convex_hull([[0, 0], [2, 0], [1, 1], [2, 2], [0, 2], [1, 3]])  # -> [[0, 0], [2, 0], [2, 2], [1, 3], [0, 2]]
convex_hull([[1, 1], [1, 1]])  # -> [[1, 1]]

另有 8 个隐藏测试和 2 个限时海量数据测试；其输入数据不公开。任务以英文提示词提供给模型。

第 1 名Mistral Devstral 2 2512已解决 · 通过 12/12 个测试生成耗时 1.3 秒 · 执行耗时 1.2 秒 · $0.0006

✓	convex_hull([[0, 0], [2, 0], [1, 1], [2, 2], [0, 2], [1, 3]])预期 [[0, 0], [2, 0], [2, 2], [1, 3], [0, 2]]	<1 毫秒
✓	convex_hull([[1, 1], [1, 1]])预期 [[1, 1]]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 981 毫秒⏱ 168 毫秒

代码

def convex_hull(points):
    points = sorted(set(tuple(p) for p in points))
    if len(points) <= 1:
        return [list(p) for p in points]

    def cross(o, a, b):
        return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])

    lower = []
    for p in points:
        while len(lower) >= 2 and cross(lower[-2], lower[-1], p) <= 0:
            lower.pop()
        lower.append(p)

    upper = []
    for p in reversed(points):
        while len(upper) >= 2 and cross(upper[-2], upper[-1], p) <= 0:
            upper.pop()
        upper.append(p)

    hull = lower[:-1] + upper[:-1]
    return [list(p) for p in hull]

第 1 名Google Gemini 3.8 Flash已解决 · 通过 12/12 个测试生成耗时 3.0 秒 · 执行耗时 1.2 秒 · $0.0016

✓	convex_hull([[0, 0], [2, 0], [1, 1], [2, 2], [0, 2], [1, 3]])预期 [[0, 0], [2, 0], [2, 2], [1, 3], [0, 2]]	<1 毫秒
✓	convex_hull([[1, 1], [1, 1]])预期 [[1, 1]]	<1 毫秒

✓ <1 毫秒✓ 8 毫秒✓ 1 毫秒✓ 1 毫秒✓ 2 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 1.0 秒⏱ 188 毫秒

代码

def convex_hull(points: list[list[int]]) -> list[list[int]]:
    pts = sorted({(p[0], p[1]) for p in points})
    if len(pts) <= 1:
        return [[p[0], p[1]] for p in pts]
    
    def cross(o, a, b):
        return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])

    lower = []
    for p in pts:
        while len(lower) >= 2 and cross(lower[-2], lower[-1], p) <= 0:
            lower.pop()
        lower.append(p)

    upper = []
    for p in reversed(pts):
        while len(upper) >= 2 and cross(upper[-2], upper[-1], p) <= 0:
            upper.pop()
        upper.append(p)

    # Check collinearity
    if len(lower) == 2 and len(upper) == 2 and lower[0] == upper[-1] and lower[-1] == upper[0]:
        # All points are collinear
        return [[pts[0][0], pts[0][1]], [pts[-1][0], pts[-1][1]]]

    hull = lower[:-1] + upper[:-1]
    return [[p[0], p[1]] for p in hull]

第 1 名MiniMax M3.1 Flash已解决 · 通过 12/12 个测试生成耗时 1.7 秒 · 执行耗时 1.2 秒 · $0.0004

✓	convex_hull([[0, 0], [2, 0], [1, 1], [2, 2], [0, 2], [1, 3]])预期 [[0, 0], [2, 0], [2, 2], [1, 3], [0, 2]]	<1 毫秒
✓	convex_hull([[1, 1], [1, 1]])预期 [[1, 1]]	<1 毫秒

✓ <1 毫秒✓ 6 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 1.1 秒⏱ 181 毫秒

代码

def convex_hull(points: list[list[int]]) -> list[list[int]]:
    pts = sorted(set((p[0], p[1]) for p in points))
    if len(pts) <= 1:
        return [list(p) for p in pts]

    def cross(o, a, b):
        return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])

    lower = []
    for p in pts:
        while len(lower) >= 2 and cross(lower[-2], lower[-1], p) <= 0:
            lower.pop()
        lower.append(p)

    upper = []
    for p in reversed(pts):
        while len(upper) >= 2 and cross(upper[-2], upper[-1], p) <= 0:
            upper.pop()
        upper.append(p)

    hull = lower[:-1] + upper[:-1]
    return [list(p) for p in hull]

第 1 名Anthropic Claude Haiku 5.5已解决 · 通过 12/12 个测试生成耗时 1.5 秒 · 执行耗时 1.3 秒 · $0.0002

✓	convex_hull([[0, 0], [2, 0], [1, 1], [2, 2], [0, 2], [1, 3]])预期 [[0, 0], [2, 0], [2, 2], [1, 3], [0, 2]]	<1 毫秒
✓	convex_hull([[1, 1], [1, 1]])预期 [[1, 1]]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 1.1 秒⏱ 176 毫秒

代码

def convex_hull(points: list[list[int]]) -> list[list[int]]:
    pts = sorted(set((p[0], p[1]) for p in points))
    n = len(pts)
    if n <= 1:
        return [[x, y] for x, y in pts]

    def cross(o, a, b):
        return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])

    lower = []
    for p in pts:
        while len(lower) >= 2 and cross(lower[-2], lower[-1], p) <= 0:
            lower.pop()
        lower.append(p)

    upper = []
    for p in reversed(pts):
        while len(upper) >= 2 and cross(upper[-2], upper[-1], p) <= 0:
            upper.pop()
        upper.append(p)

    hull = lower[:-1] + upper[:-1]
    return [[x, y] for x, y in hull]

Token 成本: $0.0028 · 轮次运行与评分规则 →